Snake Lemma

Let us introduce a useful tool for computing kernels and cokernels in a complicated diagram of modules. Although it is only marginally useful for now, it will become a major tool in homological algebra.

Snake Lemma.

Suppose we have the following diagram of A-modules and homomorphisms, where the rows are exact.

Let and for . Then we get a long exact sequence:

In diagram, the long exact sequence is drawn in green:

Proof

The proof is rather tedious so we will only prove the existence of the “snake map” .

Let so that .

Since is surjective there exists , .

Let .

Then .

Hence for a unique .

We let image of in .

This map is well-defined since if we pick another in the second step, then . Thus for some so and thus the new in the above construction is . So and have the same image in .

The rest of the proof is left as an exercise. ♦

Note

The above process is called diagram-chasing, and it is often far more effective to have it shown to you live. Here is the first part of the proof in video form.

Finitely Presented Modules

Now let us compare and for A-modules M and N. Since any A-linear map induces a B-linear map we have a map . This is clearly A-linear so it induces a B-linear map

For this to be an isomorphism, we need to introduce a new concept.

Definition.

An A-module M is finitely presented if it is finitely generated, and for some surjective map , its kernel is also finitely generated.

Note

In other words, M is finitely presented if and only if there is an exact sequence of the form

.

Intuitively, this means M can be described by a finite set of generators subjected to a finite set of relations.

Proposition 1.

If M is finitely presented, then the kernel K of any surjective map is finitely generated.

Proof

Since M is finitely presented there is an exact sequence of the form where are finite free modules. Since are free and hence projective, there exist , making the diagram commute.

By the snake lemma we have an exact sequence

.

Thus is finitely generated. Furthermore, since is finite free is finitely generated. Since and are finitely generated so is K. ♦

Exercise A

1. Find a module M over a ring A which is finitely generated but not finitely presented.

3. Prove that in a short exact sequence of A-modules , if N and P are finitely presented, so is M.

Optional Extra

The concept of finitely presented modules can be generalized further.

Definition

We say that an A-module M is of type if there is an exact sequence

where each is finite free.

Thus saying M is means it is finitely generated and saying M is means it is finitely presented. One can show more generally that if M has type then the kernel of any surjection has type . However, this is outside our scope of discussion.

This has applications in non-commutative algebra, where we study (say) left modules over a non-commutative ring and consider their cohomological properties. For details, see GTM 87, Cohomology of Groups, by Kenneth S. Brown.

Hom Functor and Induced Module

Now our main result is as follows.

Proposition 2.

If M is finitely presented and B is A-flat, we have an isomorphism

Proof

Pick an exact sequence of A-modules: . Since Hom is left-exact, for any A-module N we have an exact sequence of A-modules

And since is exact, we get an exact sequence of B-modules

Since tensor product is right-exact, we also get an exact sequence of B-modules

Again since Hom is left-exact, we get an exact sequence of B-modules

But we have natural isomorphisms

which commute with . Hence this map is an isomorphism. ♦

F.P. ⇒ (Projectivity is Local)

Since localization is exact and naturally isomorphic to , as a special case we obtain the following.

Corollary 1.

If M is finitely presented, then .

Now we are ready to prove the following result.

Proposition 3.

If M is a finitely presented A-module, then M is A-projective if and only if is -projective for all maximal .

In summary, for a finitely presented module, projectivity is a local property.

Proof

(⇒) Let M be A-projective. For a maximal ; we wish to show is -projective. Let be a surjective -linear map. Now f is A-linear, and since M is A-projective

is surjective. Since M is finitely presented and , we get a surjective:

(⇐) Suppose is -projective for all maximal . To prove M is A-projective, let be a surjective A-linear map. We need to show

is surjective; equivalently, we need to show is surjective for all maximal . But this is just